What are the four algebraic processes. Define each process and provide example please.
1. Simplify
[(x^{2}+2x-3)]/[(x+3)(x-3)]=10+y^{2}+2y-4y+8-2y^{2}
The right side requires simplification in that we need to combine like terms.
[(x^{2}+2x-3)]/[(x+3)(x-3)]=10 + y^{2 }+ 2y - 4y +8 - 2y^{2}
[(x^{2}+2x-3)]/[(x+3)(x-3)]=(-2y^{2}+y^{2})+(2y-4y)+(10+8)
[(x^{2}+2x-3)]/[(x+3)(x-3)]= - y^{2 }- 4y + 18
It is now simplified.
The right side requires simplification in that we need to combine like terms.
[(x^{2}+2x-3)]/[(x+3)(x-3)]=10 + y^{2 }+ 2y - 4y +8 - 2y^{2}
[(x^{2}+2x-3)]/[(x+3)(x-3)]=(-2y^{2}+y^{2})+(2y-4y)+(10+8)
[(x^{2}+2x-3)]/[(x+3)(x-3)]= - y^{2 }- 4y + 18
It is now simplified.
2. Factor
The numerator (x^{2} + 2x - 3) of the left side can be factored...
[(x+3)(x-1)]/[(x+3)(x-3)]=-y^{2}-4y+18
[(x+3)(x-1)]/[(x+3)(x-3)]=-y^{2}-4y+18 [The (x+3) on the numerator and the denominator cancel each other out.]
(x-1)/(x-3)=-y^{2}-4y+18
It is now factored.
[(x+3)(x-1)]/[(x+3)(x-3)]=-y^{2}-4y+18
[(x+3)(x-1)]/[(x+3)(x-3)]=-y^{2}-4y+18 [The (x+3) on the numerator and the denominator cancel each other out.]
(x-1)/(x-3)=-y^{2}-4y+18
It is now factored.
3. Evaluate
We can evaluate the fact that in its original form, the domain can include all integers except -3 and 3.
In its current form, the domain can include all integers except 3.
In its current form, the domain can include all integers except 3.
4. Solve
Graph the results.
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